ASRJC 9758 2024 Prelim P2 Solutions

[Turn Over ANDERSON SERANGOON JUNIOR COLLEGE H2 MATHEMATICS JC2 Prelim Paper 2 (100 marks) 9758 16 Sept 2024 3 hours Additional Material(s): List of Formulae (MF26) CANDIDATE NAME CLASS / READ THESE INSTRUCTIONS FIRST Write your name and class in the boxes above. Please write clearly and use capital letters. Write in dark blue or black pen. HB pencil may be used for graphs and diagrams only. Do not use staples, paper clips, glue or correction fluid. Answer all the questions and write your answers in this booklet. Do not tear out any part of this booklet. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use an approved graphing calculator. Where unsupported answers from a graphing calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. All work must be handed in at the end of the examination. If you have used any additional paper, please insert them inside this booklet. The number of marks is given in brackets [ ] at the end of each question or part question. Question number Marks 1 2 3 4 5 6 7 8 9 10 11 Total This document consists of 13 printed pages and 3 blank pages.

2 Section A: Pure Mathematics [40 marks] 1 (a) 2 Let cot d cosecd x x = =−   ( ) 22 2 22 1 d 1 1 cosec d cot 1 cot x xx + =  − +     ( ) 2 22 2 2 1 cosec d cot cosec sin 1 dcos sin =  − =−        ( ) 2sin (cos ) d −=−    OR tan sec d=−    1 cos C=− +  2 1x Cx +=− + 2 (ai) ( ) 2 1 2 2 2 2 du du 98 49 18 8 4 9 2 (3 ) u u u uu − − + = ++   21 21 1 8 3ln(4 9 ) tan2 (2)(3) 2 1 4 3ln(4 9 ) tan2 3 2 uuC uuC − −  = + − +   = + − +  (ii) 3 1 Area required = dyx ( ) 1 20 9 2 1 d49 u uuu=  ++ ( ) 21 20 11 200 18 9=d 49 982 1 d + d 49 uu uu uuu u + + −= +   1 21 0 1 4 32 ln(4 9 ) tan (from part (i))2 3 2 uuu − = + + −  

3 [Turn Over 11 12 1 4 3 1 4= 2 ln13 tan 0 ln 4 tan (0)2 3 2 2 3 1 13 4 32 ln tan units2 4 3 2 −− −     + − − + −         = + −  (b) Given 22( 2) 4( 1) 4xy+ + − = 2 2 2( 2) 4[1 ( 1) ] 2 2 1 ( 1)x y x y+ = − −  + = − − The shaded region is bounded by the section of the ellipse where x ≤ −2. Hence 22 2 1 ( 1)xy=− − − − . Volume of solid formed ( ) 2 22 1 2 2 2 1 π4 (2 1) π d 16π π 2 2 1 ( 1) d xy yy = − − = − − − − −   39.6 units (From GC)= 3 (a) 6 2 4 5 3 2 7 4 3 AB →            =−=                and 5 6 1 6 5 1 10 7 3 BE → −           = − =                Consider 4 1 3 1 2 1 15 3 5 3 3 6 2 −                = − = −                      Then ( ) ( ) ( ) 1 2 1 5 3 5 2 15 5 5 2 4 2 1 5 2 5 x y z x y z                 − =  − = − + =−                       + − + =− Thus equation of surface ABE is 5 2 5x y z− + =− (b) 5 8 13 11 6 9 1522 10 6 16 OM →            = + =                13 4 5 1115 7 12216 3 10 DM →            = − =                −

4 Line DM has equation: 45 7 1 , 3 10        = +           r  (c) Let foot of perpendicular from M to surface be N. Thus 13 1 1 15 52 16 2 ON t →        = + −          for some t 13 2 1 15 5 5 52 282 t t t  +   −  − =−  + 13 75 25 16 4 522 1 3 t t t t  + − + + + =− = Thus 13 1 23 41 15 1 15 352 3 6 52 182 3 ON →  +    = − =     +  Thus we have 41 35 26,,6 6 3N  . (d) 41 13 1 1 1 135 15 56 2 352 16 2 MN →            = − = −                1 301 25 433= + + = (e) Consider 4 5 9 1 7 8 1 3 10 13 1 +=        + =  =        +=        Since the value of  is consistent, P lines on the line DM. Also, ( ) ( )9 5 8 2 13 5− + =− Hence P also lies on the plane ABE. Thus P is the point of intersection between the line DM and the plane ABE. Let the reflection of point M about the surface ABE be the point M’. By ratio theorem (mid-point theorem), ' 2 OM OMON →→→ +=

5 [Turn Over 41 13 43 1 1 1' 2 35 15 25 3 2 652 16 56 OM ON OM → → →            = − = − =                 43 9 11 11' 25 8 2366 56 13 22 PM →            = − =−                Thus equation of the reflection of DM about the surface ABE is 9 11 8 23 , 13 22        = +           r  4 (a) Height of the cylindrical container 22(2 )ar=− 224ar=− Volume of cylindrical container V 2 2 24r a r=− Need to see that 22(2 )ar − gives the height of the container (b) 2 2 24V r a r=− ( ) ( ) ( ) 2 2 2 1 2 2 2 2 2 2 dd 4dd d1 2 4 4 2d2 V r a rrr V r a r r a r rr   − =− = − + − −  3 22 22 d 24d 4 Vr r a rr ar = − − − ( ) 2 2 3 22 24d d 4 r a r rV r ar  −− = − ( ) 22 22 83d d 4 r a rV r ar − = −  ( ) ( ) 2 2 2 80 4 50 3600 60 0 r r rr =− = = 2a cm r cm

6 ( ) ( ) ( ) 22 22 (60) 8 50 3 60d 6900d 4 50 60 V r  −== −   d d d d d d d100 6900 d V V r t r t r t = = 1d1 cm sd 69 r t −= (c) ( )( ) 22 2 2 3 2 2 3d d 4 r a r a rV r ar +− = −  ( )( ) 22 d 0d 2 2 3 2 2 3 0 4 V r r a r a r ar  = +− = − 2 2 2 20 (rej 0) or (rej 0) or 33 aar r r r r=  =−  = ( ) ( ) 22 22 22When 0 < , 3 2 2 3 d0 & 2 2 3 0 0. d4 22When , 3 2 2 3 d0 & 2 2 3 0 0. d4 ra r a r Var rar ra r a r Var rar  +  −    −  +  −    −   22 3 a −   22 3 a  22 3 a +   d d V r +ve 0 – ve Slope V is maximum when 2 2 2 6 33 aar== When 26 3 ar= , 22 2 33 3 2 6 2 6 433 16 16 3 cm933 aaVa aaV      =−           ==

7 [Turn Over (d) Section B: Probability and Statistics [60 marks] 5 (i) Required probability 2 4 3 2 1 2 55 + + +== (ii) Required probability 7 3 5 2 7 3 5 2 3 1 2 1 3 1 2 1 4 3 4 3 4! 3! 4! 3! 2! 2! 2! 2! 27 5 7 5 C C C C C C C C      = + −          = 0.50099 =0.501 (3 s.f) 6 (i) Table of outcomes: 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 4 6 3 0 3 6 9 The probability distribution of T is given by: T 0 1 2 3 4 6 9 P(T = t) 7 16 1 16 2 16 1 8= 2 16 1 8= 1 16 2 16 1 8= 1 16 (ii) ( ) 7 1 1 1 1 1 1 9E 0 1 2 3 4 6 916 16 8 8 16 8 16 4T              = + + + + + + =                          

8 ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 22 7 1 1 1 1 1 1 49E 0 1 2 3 4 6 916 16 8 8 16 8 16 4 49 9 115Var E E 4 4 16 T T T T              = + + + + + + =                           = − = − =   ( ) ( ) ( ) ( ) ( ) ( ) P 2 P 2 P 2 P 2 P 2 T T T TT −  = −  + − − =  + +  − iii           = ( ) ( )P 7.18095 P 1.81905TT +  = ( ) ( ) ( )P 9 P 0 +P 1T T T= + = = 1 7 1 9 16 16 16 16= + + = or 0.5625 7 (a) Let X be the mass, in grams, of a randomly chosen packet of semolina. X ~ 2N(225, 25 ) 224 N(4 225, 4 25 )X  ( ) 24 N 900, 100X P(850 4 1050) 0.62466 (5 sf)